
\prob{000C}{单位等边三角形}

\begin{figure}[htbp]
  \centering
  \image{000C}
  \caption{000C：单位等边三角形} \label{fig:000C}
\end{figure}

如图~\ref{fig:000C}，$\triangle ABC$是等边三角形，$AB = AC = BC = 1$，$P$在线段$AB$上，$Q$在$BC$的延长线上且$AP = CQ$，$PE \perp AC$于$E$，求$DE$的长。
\problabels{yellow/平面几何, green/长度问题}

\ans{$DE = 1/2$}

\subsection{构造8形全等} \label{subsec:000C-8eq}

\begin{figure}[htbp]
  \centering
  \image{000C-8eq}
  \caption{\nameref{subsec:000C-8eq}：通过构造8形全等和一个等边三角形求解。}
  \label{fig:000C-8eq}
\end{figure}

基本思路：通过作一个等边三角形构造8形全等，进而求出$DE$的长。

如图~\ref{fig:000C-8eq}，作$PC' \parallel BQ$，交$AC$于点$C'$。

\begin{align*}
  \because  {}& PC' \parallel BQ \\
  \therefore{}& \angle APC' = \angle B, \angle AC'P = \angle ACB, \\
  & \angle C'PD = \angle Q \\
  \because  {}& \text{$\triangle ABC$ 是等边三角形} \\
  \therefore{}& \angle B = \angle ACB = 60^\circ \\
  \therefore{}& \angle APC' = \angle AC'P = 60^\circ \\
  \therefore{}& \text{$\triangle APC'$ 是等边三角形} \\
  \therefore{}& \angle A = \angle AC'P, AP = C'P \\
  \because  {}& PE \perp AC \\
  \therefore{}& \angle AEP = \angle C'EP = 90^\circ \\
  \because  {}& \text{在 $\triangle AEP$ 与 $\triangle C'EP$ 中} \\
  & \begin{cases}
    \angle A = \angle PC'E \\
    \angle AEP = \angle C'EP \\
    EP = EP \\
  \end{cases} \\
  \therefore{}& \triangle AEP \cong \triangle C'EP \\
  \therefore{}& AE = C'E \\
  \text{又}&\because AP = CQ \\
  \therefore{}& C'P = CQ \\
  \because  {}& \text{在 $\triangle C'DP$ 与 $\triangle CDQ$ 中} \\
  & \begin{cases}
    \angle C'DP = \angle CDQ \\
    \angle C'PD = \angle Q \\
    C'P = CQ \\
  \end{cases} \\
  \therefore{}& \triangle C'DP \cong \triangle CDQ \\
  \therefore{}& CD = C'D \\
  \because  {}& DE = C'D + C'E \\
  \therefore{}& DE = CD + AE \\
  \because  {}& DE + CD + AE = AC = 1 \\
  \therefore{}& 2DE = 1 \\
  \therefore{}& DE = \frac12 \\
\end{align*}

综上，$DE$的长度为$1/2$。
